What do we mean by saying "VC dimension gives a LOOSE, not TIGHT bound"?
But, the literature (i.e. Learning from Data) states that VC gives a loose bound and that in real applications, learning models with lower VC dimension tend to generalize better than those with higher VC dimension.


It's true that people often use techniques such as regularisation to avoid over-parametrized models. However, I think it's dangerous to say that, in real applications, those models are really generalizing better, given that you're typically assessing their generalization ability by using a finite validation dataset (possibly selected in a biased way). Moreover, note that the validation dataset is often ignored in certain bounds and the only thing that is taken into account is the expected risk and the empirical risk (on the training data).

In any case, the $\mathcal{VC}$ bounds may be "loose" because


the $\mathcal{VC}$ dimension is often expressed with the big-$\mathcal{O}$ notation (e.g. see this answer)
the bounds often involve probabilities and uncertainties (e.g. see this other answer)
there aren't stricter bounds (i.e. no one found better bounds yet)



  Is the VC generalization bound loose due to its universality? Meaning, its results apply to all hypothesis sets, learning algorithms, input spaces, probability distributions, and binary target functions.


I think the answer is yes. In fact, often, the specific tasks or learning algorithms are ignored in the analyses, but, in general, you may exploit your knowledge of the problem to improve e.g. the number of useful examples that the learning algorithm could use.